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Homework Help: How do i find the value of p

  1. Apr 17, 2009 #1
    how do i find the value of p so that the following integral converges

    integral (from 0 to infinity)

    [tex]\int[/tex]arctan(x)/xp dx

    actually integrating this looks almost impossible, if not impossible, if you know how please show me,
    if not, how else can i find the value from p??
    Last edited: Apr 17, 2009
  2. jcsd
  3. Apr 17, 2009 #2
    arctan tends to a constant (pi/2) for x to infinity. So, you would expect that p > 1. You can prove this by doing a partial integration. Assume that p > 1 and integrate the 1/x^p factor. You can ten easily prove that the integral converges. Then, using the same partial integration method, you prove that for p = 1 the integral diverges.

    Finally you can use that for p < 1 the integrand is always larger than what it is for p = 1 for any value of x, to prove that for p < 1 the integral also diverges.
  4. Apr 17, 2009 #3
    that seems very reasonable, however the correct answer is (supposed to be)

    i think that this comes from the fact that there could be divergance due to what happens at the bottom limit (when x=0, we are dividing by 0) but i am not sure.
  5. Apr 17, 2009 #4
    Yes, but then it is the integral from zero to infinity, not from 1 to infinity. Note that this is up to trivial transformations the same integral you posted before, i.e. the integral of tan^p(x) from zero to pi/2.

    The integral can be computed exactly, if you are interested I can explain how to do that.
  6. Apr 17, 2009 #5
    my mistake, the correct question states from 0 to infinity, i'll change it in my post,
    how do i get to 1 < p < 2 though
  7. Apr 17, 2009 #6
    also, how would you integrate?
  8. Apr 17, 2009 #7
    Split the integral in an integral from zero to 1 and an integral from 1 to infinity. Then prove that the first integral converges for p < 2.
  9. Apr 17, 2009 #8
    thanks, but the question is HOW to prove it?
  10. Apr 17, 2009 #9
    You can use that arctan(x) < x. Postpone proving this and proceed with the proof first. Then because the integral from zero to 1 of
    arctan(x)/x^p dx will be less than the integral of x/x^p dx = integral of dx/x^(p-1) and this integral converges for p < 2 you have the result that the integral of arctan(x)/x^p will converge for p < 2.

    However, this does not prove that the integral won't also converge for p equal to 2 or even larger values of p. So, what you do is you consider the special case of p = 2 and then show that the integral diverges. You can do that using a partial integration:

    Integral of arctan(x)/x^2 dx =

    -[1/x arctan(x)] + Integral of 1/x 1/(1+x^2) dx

    To be fully rigorous take the lower limit to be epsilon. The purpose is then to show that the limit to epsilon is zero does not exist. The limit of arctan(x)/x for x to zero does exist, so we only need to consider the last term. Now, if you just do a partial fraction expansion and isolate the 1/x, you are done.

    Finally, you say that for p > 2 the integrand is larger than what it is for p =2 for any value of x. So, given that for p = 2 the integral diverges, it will also diverge for p larger than 2.
  11. Apr 18, 2009 #10
    how do i integrate 1/x 1/(1+x^2) dx to prove that p=2 diverges, i tried integration in parts but doesnt really help. how did you do it?
  12. Apr 18, 2009 #11
    Partial fractions:

    1/x 1/(1+x^2) = A/x + (C + D x)/(1+x^2)

    C = 0 because the left hand side is an odd function:

    1/x 1/(1+x^2) = A/x + (C + D x)/(1+x^2)

    The A/x term can be found by multiplying both sides by x and taking the limit x to zero. This gives A = 1. We don't need to compute D, we just need that:

    1/x 1/(1+x^2) = 1/x + term proportional to x/(1+x^2)

    The last term can be integrated from zero to 1, it does not diverge. The fist term integrated from epsilon to 1 yileds a Log(epsilon) term and the limit to epsilon to zero does not exist, so the integral diverges.

    Note that it is essential in this argument to consider both terms and to show that the second term does converge. The reason is that singularites can sometimes cancel. E.g. the integral of x from zero to 1 clearly exists. But, if some mathematical manipulation leads to the x being written as:

    1/x + (x - 1/x)

    and you were to say tat this is not convergent because the first term diverges when integrated from zero to 1, then that's clearly wrong. The second term also diverges and, in this case, the sum of the two does converge.
  13. Apr 18, 2009 #12
    also i understand everything youve done, except for the part where you check convergence for p=1, what do you mean by partial integration? i tried by integrationg arctgx/x and got stuck with that integration too. why are you integrating only1/x^p? is it because arctgx/x is similar to (pi/2)/x for large x? is it okay to only look at the part where x->inf, and not take into account the rest?
  14. Apr 18, 2009 #13
    p = 1 is where the integral of 1/x^p diverges. As long as 1/x^p converges you have a proof that arctan(x)/x^p also converges because it is less than pi/2 1/x^p . This is why for p = 1 you need to look at arctan(x)/x^p.

    You can do a partial integration:

    Int atn(x)/x dx = Log(x) atn(x) - Int Log(x)/(1+x^2) dx

    The last integral from 1 to infinity converges, the first term does not converge at infinity. This then proves that for p = 1 you know that the integral from 1 to infinity does not converge.

    Note that the integral of Log(x)/(1+x^2) converges because for arbitrary c you have that for large enough x you have Log(x) < A x^c. So, you can take c = 0.001, replace the numerator by x^2 and end up with an integrand that is larger and the integral still converges.
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